A man pulls a sled at a constant velocity across a horizontal snow surface. if a force of 80 n is being applied to the sled rope
2 answers:
Answer:
The force of friction between sled and snow is 48.14 N
Explanation:
It is given that,
Force applied on the sled, F = 80 N
The angle with the sled and the ground is 53 degrees. We need to find the force of friction between sled and snow.
The horizontal force acting in the horizontal direction is given by :
So, the force of friction between sled and snow is 48.14 N. Hence, this is the required solution.
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Answer:Orbital period =21.22hrs
Explanation:
given that
mass of earth M = 5.97 x 10^24 kg
radius of a satellite's orbit, R= earth's radius + height of the satellite
6.38X 10^6 + 3.25 X10^7 m =3.89 X 10^7m
Speed of satellite, v=
where G = 6.673 x 10-11 N m2/kg2
V= \sqrt (6.673x10^-11 x 5.97x10^ 24)/(3.89 X 10^ 7m)
V =10,241082.2
v= 3,200.2m/s
a) Orbital period
=
V=
T= 2 r/ V
= 2 X 3.142 X 3.89 X 10^7m/ 3,200.2m/s
=76,385.1 s
60 sec= 1min
60mins = 1hr
76,385.1s =hr
76,385.1/3600=21.22hrs
Answer:
Magnitude the net torque about its axis of rotation is 1.3338 Nm
Explanation:
The radius of the wrapped rope around the drum, r = 1.24 m
Force applied to the right side of the drum, F = 4.56 N
The radius of the rope wrapped around the core, r' = 0.57 m
Force on the cylinder in the downward direction, F' = 7.58 N
Now, the magnitude of the net torque is given by:
where
= Torque due to Force, F
= Torque due to Force, F'
Now,
The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.
Now, the magnitude of the net torque: